Week 6 Questions On Confidence Intervals: 1. On a busy Sunday morning, a waitress randomly sampled customers about their preference for morning beverages, Specifically, she wanted to find out how many people preferred coffee over tea. The proportion of customers that preferred coffee was 0.42 with a margin of error 0.07. Construct a confidence interval for the proportion of customers that preferred coffee.

(0.42 – 0.07), (0.42 + 0.07) = (0.35), (0.49)

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2. A company sells juice in 1quart bottles. In a quality control test, the company found the mean volume of juice in a random sample of bottles was X = 31 ounces, with a marginal error of 3 ounces. Construct a confidence interval for the mean number of ounces of juice bottled by this company.

(31-3), 31+3) = (28), (34)

3. Randomly selected employees at an office were asked to take part in a survey about overtime. The office manager wanted to find out how many employees worked overtime in the last week. The proportion of employees that worked overtime was 0.83, with a margin of error of 0.11.

(0.83 – 0.11), (0.83 + 0.11) = (0.72), (0.94)

4. A random sample or garter snakes were measured, and the proportion of snakes that were longer than 20 inches in length recorded. The measurements resulted in a sample proportion of p = 0.25 with a sampling standard deviation of Op = 0.05. Write a 68% confidence interval for the true proportion of garter snakes that were over 20 inches in length.

(.25 – .05). (.25 + .05) = (0.20), (.30)

5. The average number of onions needed to make French onion soup from the population of recipes is unknown. A random sample of recipes yields a sample mean of x = 8.2 onions. Assume the sampling distribution of the mean has a standard deviation of 2.3 onions. Use the Empirical Rule to construct a 95% confidence interval for the true population mean number of onions. Since 95% falls in 2 SD’s the calculation would be (8.2 – 4.6) “4.6 is the margin of error”, (8.2 + 4.6) = (3.6) , (12.8)

6. In a survey, a random sample of adults was asked whether a tomato is a fruit or vegetable. The survey resulted in a sample proportion of 0.58 with a sampling standard deviation of 0.08 who stated a tomato is a fruit. Write a 99.7 confidence interval for the true proportion of the number of adults who stated the tomato is a fruit.

(0.58 – 3 x 0.08), (0.58 + 3 x 0.08) =(0.58 – .24), (0.58 + .24) = 0.34 + 0.82

7. A college admissions director wishes to estimate the mean number of students currently enrolled. The age of a random sample of 23 students is given below. Assume the ages are approximately normally distributed. Use Excel to construct a 90% confidence interval for the population mean age. Round your answer to 2 decimal places and use increasing order. Use the week 6 worksheet to get mean and SD.

Data

25.8 Mean 23.1043

22.2 Sample Standard Deviation

1.3693

22.5

22.8

24.6

24

22.6

23.6

22.8

23.1

21.5

21.4

22.5

24.5

21.5

22.5

20.5

23

25.1

25.2

23.8

21.8

24.1

Confidenc

e

Level 0.900

n 32

Mean 23.1043

StDev 1.3693

pop stdev no

SE

0.242060

t 1.696

Margin of

Error 0.410534

Lower

Limit

22.693766

Upper

Limit

23.514834

Lower margin of error = 22.69 and upper limit is 23.51

8. Suppose that the scores of bowlers in a particular league follow a normal distribution such that a standard deviation of the population is 12. Find the 95% confidence interval of the mean score for all bowlers in this league using the accompanying data set of 40 random scores. Round your answers to 2 decimal places using ascending order.

Lower Limit = 90.78 Upper Limit = 98.22

Confidenc

e

Level 0.950

n 40

Mean 94.5000

StDev 12.0000

pop stdev yes

SE

1.897367

z 1.960

Margin of

Error 3.718839

Lower

Limit

90.781161

Upper

Limit

98.218839

9. In the survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for the proportion of adults who regularly lie to people conducting surveys. Use excel to create the confidence interval rounding to 4 decimal places.

Lower Limit = 0.1238 Upper Limit = 0.2012

Confidence

Level

0.990

n 603

Number of

Successes 98

Sample

Proportion

0.162521

SE 0.015024

z 2.576

Margin of Error 0.038702

Lower Limit 0.123819

Upper Limit 0.201222

10. In a random sampling of 350 attendees at a minor league baseball game, 184 said that they bought food the concession stand. Create a 95%confidence interval for the proportion of fans who bought food from the concession stand. Use excel to create the confidence interval rounding to 4 decimal places.

Lower limit = 0.4734 Upper Limit = 0.5780

Confidence

Level

0.950

n 350

Number of

Successes 184

Sample

Proportion

0.525714

SE 0.026691

z 1.960

Margin of Error 0.052314

Lower Limit 0.473400

Upper Limit 0.578028

11. Suppose that the weight of tight ends in a football league are normally distributed such that sigma squared = 1,369. A sample of 49 tight ends was randomly selected and the weights are given in the table below. Use Excel to create a 95% confidence interval for the mean weight of the tight ends in this league. Rounding your answers to 2 decimal places and using ascending order. (Have to get a square root of 1369 which is 37). The population sample is yes.

Lower limit = 241.42 Upper Limit = 262.14

Confidence 0.950

Level

n 49

Mean 251.7755

StDev 37.0000

pop stdev yes

SE

5.285714

z 1.960

Margin of

Error 10.360000

Lower

Limit

241.415500

Upper

Limit

262.135500

12. Suppose heights, in inches of orangutans, are normally distributed and have a known population standard deviation of 4 inches. A random sample of 16 orangutans is taken and gives a sample mean of 56 inches. Find the confidence interval of the population means with a 95% confidence level.

Lower limit = 54.04 and Upper Limit = 57.96

13. The population standard deviation for the total snowfalls per year in a city is 13 inches. If we want to be 95% confident that the sample mean is within 3 inches of the true population mean, what is the minimum sample size that should be taken?

Answer: 73 snowfalls

Minimum Sample Size μ for the population mean

Confidence Level 0.950

StDev 13

Error 3

z-Value 1.960

Minimum Sample

Size

73

14. The population standard deviation for the body weights of employees of a company is 10 pounds. If we want to be 95% confident that the sample mean is within 3 pounds of the true population mean, what is the minimum sample size that should be taken.

Answer: 43 employees

Minimum Sample Size μ for the population mean

Confidence Level 0.950

StDev 10

Error 3

z-Value 1.960

Minimum Sample

Size

43

15. The length, in words, of the essays written for a contest, is normally distributed with a population standard deviation of 442 words and an unknown population mean. If a random sample of 24 essays is taken and results in a sample mean of 1330 words, find a 99% confidence interval for the population mean. Round to

two decimal places.

Answer: Lower limit = 1097.59 upper Limit = 1562.41

16. Brenda wants to estimate the percentage of people who eat fast food at least once per week. She wants to create a 95% Confidence interval which has an error bound of at most 2%. How many people should be polled to create the confidence interval?

Answer: 2401

Minimum Sample Size p for Proportion

Confidence Level 0.95

0 Enter decimal

Sample Proportion 0.5 If the sample proportion is unknown enter

0.5

Error 0.02 Write percentage as a decimal

z-Value 1.96

0

Minimum Sample

Size

2401

17. Suppose a clothing store wants to determine the current percentage of customers who are over the age of forty. How many customers should the company survey in order to be 92% confident that the estimated (sample) proportion is within 5% of the true population proportion of customers who are over the age of 40?

Answer: 307

Minimum Sample Size p for Proportion

Confidence

Level

0.920 Enter decimal

Sample

Proportion

0.5 If sample proportion is unknown enter

0.5

Error 0.05 Write percentage as a decimal

z-Value 1.751

Minimum Sample Size 307

18. Suppose the scores of a standardized test are normally distributed. If the population standard deviation is 2 points, what minimum sample size is needed to be 90% confident that the sample mean is within 1 point of the true population mean? Be sure to round up to the nearest integer.

Provide your answer below: 11

Minimum Sample Size μ for the population mean

Confidence Level 0.900

StDev 2

Error 1

z-Value 1.645

Minimum Sample

Size

11

19. The number of square feet per house is normally distributed with a population standard deviation of 197 square feet and an unknown population mean. If a random sample of 25 houses is taken and results in a sample mean of 1820 square feet, find a 99% confidence interval for the population mean. Round to 2 decimal places.

Answer: 1718.51 – 1921.49

t or z Confidence Interval for µ

Confidence

Level

0.990

n 25

Mean 1,820.0000

StDev 197.0000

pop stdev yes

SE

39.400000

z 2.576

Margin of Error 101.494400

Lower Limit 1718.505600

Upper Limit 1921.494400

20. Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 6 points and an unknown population mean. A random sample of 22 scores is taken and gives a sample mean of 92 points. Identify the parameters needed to calculate a confidence interval at the 98% confidence level. Then find the confidence interval.

x = 92

σ = 6

n = 22

zα2= 2.326

(89.02, 94.98)

t or z Confidence Interval for µ

Confidence

Level

0.980

n 22

Mean 92.0000

StDev 6.0000

pop stdev yes

SE

1.279204

z 2.326

Margin of Error 2.975429

Lower Limit 89.024571

Upper Limit 94.975429

21. Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 6 points and an unknown population mean. A random sample of 22 scores is taken and gives a sample mean of 92 points. What is the correct interpretation of the 95% confidence interval? We can estimate that 98% of the time the test is taken, a student scores between 89.02 and 94.98 points. We can estimate with 98% confidence that the true population mean score is between 89.02 and 94.98 points. We can estimate with 98% confidence that the sample mean score is between 89.02 and 94.98 points

22. The weights of running shoes are normally distributed with a population standard deviation of 3 ounces and an unknown population mean. If a random sample of 23 running shoes is taken and results in a sample mean of 18 ounces, find a 90%confidence interval for the population mean. Round the final answer to two decimal places.

Answer: 16.97 – 19.03

23. The germination periods, in days, for grass seed are normally distributed with a population standard deviation of 5 days and an unknown population mean. If a random sample of 17 types of grass seed is taken and results in a sample mean of 52days, find a 80% confidence interval for the population mean.

Select the correct answer below:

(50.45,53.55)

(50.01,53.99)

(49.85,54.15)

(49.62,54.38)

(49.18,54.82)

(48.88,55.12)

24. The speeds of vehicles traveling on a highway are normally distributed with a population standard deviation of 7 miles per hour and an unknown population mean. If a random sample of 20 vehicles is taken and results in a sample mean of 60miles per hour, find a 98% confidence interval for the population mean.

Round the final answer to two decimal places.

Answer 56.36 – 63.64

t or z Confidence Interval for µ

Confidence

Level

0.980

n 20

Mean 60.0000

StDev 7.0000

pop stdev yes

SE

1.565248

z 2.326

Margin of Error 3.640766

Lower Limit 56.359234

Upper Limit 63.640766

25. Suppose finishing times for cyclists in a race are normally distributed and have a known population standard deviation of 6minutes and an unknown population mean. A random sample of 18 cyclists is taken and gives a sample mean of 146minutes. Find the confidence interval for the population means with a 99% confidence level.

Answer: 142.36 – 149.64

t or z Confidence Interval for µ

Confidence

Level

0.990

n 18

Mean 146.0000

StDev 6.0000

pop stdev yes

SE 1.414214

z 2.576

Margin of Error 3.643014

Lower Limit 142.356986

Upper Limit 149.643014

26. Suppose the germination periods, in days, for grass seed are normally distributed. If the population standard deviation is 3days, what minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean?

Answer: 25 seeds

Minimum Sample Size μ for the population mean

Confidence Level 0.900

StDev 3

Error 1

z-Value 1.645

Minimum Sample

Size

25

27. Suppose the number of square feet per house is normally distributed. If the population standard deviation is 155 square feet, what minimum sample size is needed to be 90% confident that the sample mean is within 47 square feet of the true population mean?

Answer: 30 houses

Minimum Sample Size μ for population mean

Confidence Level 0.900

StDev 155

Error 47

z-Value 1.645

Minimum Sample

Size

30

28. In a survey of 1,000 adults in a country, 722 said that they had eaten fast food at least once in the past month. Create a 95% confidence interval for the population proportion of adults who ate fast food at least once in the past month. Use Excel to create the confidence interval, rounding to four decimal places.

Answer: 0.6942 – 0.7498

Confidence Interval for p

Proportions

Confidence Level 0.950

n 1000

Number of

Successes

722

Sample Proportion 0.722000

SE 0.014167

z 1.960

Margin of Error 0.027768

Lower Limit

0.694232

Upper Limit 0.749768

29. A college admissions director wishes to estimate the mean age of all students currently enrolled. The age of a random sample of 23 students is given below. Assume the ages are approximately normally distributed. Use Excel to construct a 90% confidence interval for the population mean age. Round your answers to two decimal places and use increasing order.

Answer: 22.61 – 23.59

Data

25.8 Mean 23.1043

22.2 Sample Standard Deviation 1.3693

22.5

22.8

24.6

24

22.6

23.594536

0.900

23

23.1043

1.3693

no

0.285519

1.717

0.490236

22.614064

23.6

22.8

23.1

21.5

21.4

22.5

24.5

21.5

22.5

20.5

23

25.1

25.2

23.8

21.8

24.1

t or z Confidence Interval for µ

Confidence

Level n Mean

StDev

pop stdev

SE t Margin

of Error

Lower Limit

Upper Limit

30. The yearly incomes, in thousands, for 24 random married couples living in a city are given below. Assume the yearly incomes are approximately normally distributed. Use Excel to find the 95% confidence interval for the true mean, in thousands. Round your answers to three decimal places and use increasing order.

Answer: 58.984 – 59.026

Data

59.015 Mean 59.0050

58.962 Sample Standard Deviation 0.0494

58.935

58.989

58.997

58.97

59

59.014

59.001

59.003

58.992

58.926

59.032

58.958

59.093

58.955

59.003

58.952

59.057

59.056

59.074

59.128

59.001

59.007

t or z Confidence Interval for µ

Confidence

Level

0.950

n 24

Mean 59.0050

StDev 0.0494

pop stdev no

SE 0.010084

t 2.069

Margin of

Error

0.020863

Lower Limit 58.984137

Upper Limit 59.025863

31. A tax assessor wants to assess the mean property tax bill for all homeowners in a certain state. From a survey ten years ago, a sample of 28 property tax bills is given below. Assume the property tax bills are approximately normally distributed. Use Excel to construct a 95% confidence interval for the population mean property tax bill. Round your answers to two decimal places and use increasing order.

Answer: 1185.91 – 1595.59

32. The table below provides a random sample of 20 exam scores for a large geology class. Use Excel to construct a 90% confidence interval for the mean exam score of the class. Round your answers to one decimal place and use ascending order.

Answer: 79.7 – 88.5

t or z Confidence Interval for µ

Confidence

Level

0.900

n 20

Mean 84.1000

StDev 11.4200

pop stdev no

SE 2.553590

t 1.729

Margin of

Error

4.415156

Lower Limit 79.684844

Upper Limit 88.515156

33. Suppose scores on exams in statistics are normally distributed with an unknown population mean. A sample of 26 scores is given below. Use Excel to find a 90% confidence interval for the true mean of statistics exam scores. Round your answers to one decimal place and use increasing order.

Answer: 67.2 – 69.4

Confidenc

e

Level 0.900

n 26

Mean 68.3077

StDev 3.2220

pop stdev no

SE

0.631886

t 1.708

Margin of

Error 1.079262

Lower

Limit

67.228438

Upper

Limit

69.386962

33. In a city, 22 coffee shops are randomly selected, and the temperature of the coffee sold at each shop is noted. Use Excel to find the 90% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed. Round your answers to two decimal places and use increasing order.

Answer: 153.21 – 161.43

t or z Confidence Interval for µ

Confidence

Level

0.900

n 22

Mean 157.3182

StDev 11.2054

pop stdev no

SE

2.388999

t 1.721

Margin of Error 4.111468

Lower Limit 153.206732

2.160415

2.485

5.368632

66.496768

77.234032

0.980

26

71.8654

11.0160

no

26 is given

Upper

Limit

161.429668

34. Weights, in pounds, of ten-year-old girls are collected from a neighborhood. A sample of below. Assuming normality, use Excel to find the 98% confidence interval for the population mean weight μ. Round your answers to three decimal places and use increasing order.

Answer: 66.497 – 77.234

t or z Confidence Interval for µ

Confidence

Level n Mean

StDev pop

stdev

SE t Margin

of Error

Lower Limit

Upper Limit

35. A sample of 22 test tubes tested for a number of times they can be heated on a Bunsen burner before they crack is given below. Assume the counts are normally distributed. Use Excel to construct a 99% confidence interval for μ. Round your answers to two decimal places and use increasing order.

Answer: 1071.77 – 1477.33

36. The monthly incomes from a random sample of 20 workers in a factory is given below in dollars. Assume the population has a normal distribution and has a standard deviation of $518. Compute a 98% confidence interval for the mean of the population. Round your answers to the nearest dollar and use ascending order.

Answer: 11,833 – 12,372

37. Assume the distribution of commute times to a major city follows the normal probability distribution and the standard deviation is 4.5 minutes. A random sample of 104 commute times is given below in minutes. Use Excel to find the 98%confidence interval for the mean travel time in minutes. Round your answers to one decimal place and use ascending order.

Answer: 25.9 – 27.9

38. Installation of certain hardware takes a random amount of time with a standard deviation of 7 minutes. A computer technician installs this hardware on 50 different computers. These times are given in the accompanying dataset. Compute a 95% confidence interval for the mean installation time. Round your answers to two decimal places and use ascending order.

Answer: 40.76 – 44.64

Confidence

e

Level 0.950

n 50 39. Assume that farm sizes in a particular region are normally

Mean 42.7000

StDev 7.0000

distributed with a population standard deviation of 150 acres. A pop stdev y random sample of 50 farm sizes in this region is given below in

acres. Estimate the mean farm size for this region SE with 90%confidence. Round your answers to two decimal

0.989949

z 1.960 places and use ascending order.

Margin of

Answer: 474.87 – 544.65

Error 1.940301

Lower

Limit

40.759699

Upper

Limit

44.640301

40. The amount of time that customers stay in a certain restaurant for lunch is normally distributed with a standard deviation of 17 minutes. A random sample of 50 lunch customers was taken at this restaurant. Construct a 99% confidence interval for the true average amount of time customers spend in the restaurant for lunch. Round your answers to two decimal places and use ascending order.

Answer: 44.89 – 57.27

41. Recent studies have shown that out of 1,000 children, 885 children like ice cream. What is the 99% confidence interval for the true proportion of children who like ice cream, based on this sample? Round z⋆ to two decimal places and other answers to four decimal places. Provide your answer below: .8590 – .9110 Confidence Interval for p Proportions

Confidence Level 0.990 Enter decimal

n 1000

Number of

Successes

885

Sample Proportion 0.885000

SE 0.010088

z 2.576

Margin of Error 0.025988

Lower Limit 0.859012

Upper Limit 0.910988

0.393393

0.026770

1.960

0.052469

0.340925

0.445862

Sample Proportion

SE

z Margin of

Error

Lower

Limit

Upper

Limit

0.950

333

131

Confidence

Level n Number of

Successes

42. A large company is concerned about the commute times of its employees. 333 employees were surveyed, and 131employees said that they had a daily commute longer than 30 minutes. Create a 95% confidence interval for the proportion of employees who have a daily commute longer than 30 minutes. Use Excel to create the confidence interval, rounding to four decimal places.

Provide your answer below: .3409 –

.4459

Confidence

Interval for p

Proportions

43. The following data represent a random sample for the ages of 41 players in a baseball league. Assume that the population is normally distributed with a standard deviation of 2.1 years. Use Excel to find the 98% confidence interval for the true mean age of players in this league. Round your answers to three decimal places and use ascending order.

Answer: 27.579 – 29.104

t or z Confidence Interval for µ

Confidence

Level

0.980

n 41

Mean 28.3415

StDev 2.1000

pop stdev yes

SE 0.327965

z 2.326

Margin of Error 0.762846

Lower Limit 27.578654

Upper Limit 29.104346

44. In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample of 25 pieces of carry-on luggage was collected and weighed in pounds. Assume that the population is normally distributed with a standard deviation of 5 pounds. Find the 95% confidence interval of the mean weight in pounds. Round your answers to two decimal places and use ascending order.

Answer: 15.36 – 19.28

t or z Confidence Interval for µ

Confidence

Level

0.950

n 25

Mean 17.3200

StDev 5.0000

pop stdev yes

SE 1.000000

z 1.960

Margin of Error 1.960000

Lower Limit 15.360000

Upper Limit 19.280000

45. A company wants to determine a confidence interval for the average CPU time of its teleprocessing transactions. A sample of 70 random transactions in milliseconds is given below. Assume that the transaction times follow a normal distribution with a standard deviation of 600 milliseconds. Use Excel to determine a 98% confidence interval for the average CPU time in milliseconds. Round your answers to the nearest integer and use ascending order.

Answer: 5907 – 6240 t or z Confidence Interval for µ

Confidence

Level

0.980

n 70

Mean 6,073.4286

StDev 600.0000

pop stdev yes

SE 71.713717

z 2.326

Margin of Error 166.806105

Lower Limit 5906.622495

Upper Limit 6240.234705

46. The number of hours worked per year per adult in a state is normally distributed with a standard deviation of 37. A sample of 115 adults is selected at random, and the number of hours worked per year per adult is given below. Use Excel to calculate the 98% confidence interval for the mean hours worked per year for adults in this state. Round your answers to two decimal places and use ascending order.

Answer: 2090.03 – 2106.09

47. An automobile shop manager timed 27 employees and recorded the time, in minutes, it took them to change a water pump. Assuming normality, use Excel to find the 99% confidence interval for the true mean. Round your answers to three decimal places and use increasing order.

Answer: 15.499 – 19.139

Confidence

Level n Mean

StDev

pop stdev

SE

t Margin of

Error

Confidence

Level n Mean

StDev

pop

stdev

SE

t Margin of

Error

Lower Limit

Upper Limit

Confidence

Level n Mean

StDev

pop stdev

SE0.645380

t2.064

Margin of

Error

1.332064

Lower

Limit

79.951936

Upper

Limit

82.616064

49. The heart rates for a group of 21 students taking a final exam are given below. Assume the heart rates are normally distributed. Use Excel to find the 95% confidence interval for the true mean. Round your answers to two decimal places and use increasing order.

Answer: 91.31 – 95.17

t or z Confidence Interval for µ

0.950

25

81.2840

3.2269

no

Answer: 79.95 – 82.62

t or z Confidence Interval for µ

48. A type of golf ball is tested by dropping it onto a hard surface from a height of 1 meter. The height it bounces is known to be normally: distributed. A sample of 25 balls is tested and the bounce heights are given below. Use Excel to find a 95%confidence interval for the mean bounce height of the golf ball. Round your answers to two decimal places and use increasing order.

0.654888

2.779

1.819935

15.498565

19.138435

0.990

27

17.3185

3.4029

no

t or z Confidence Interval for µ

0.950

21

93.2381

4.2415

no

0.925571

2.086

1.930741

Lower

Limit

91.307359

Upper

Limit

95.168841

50. Suppose a clothing store wants to determine the current percentage of customers who are over the age of forty. How customers should the company survey in order to be 90% confident that the estimated (sample) proportion is within 4percentage points of the true population proportion of customers who are over the age of forty?

Answer: 423

Minimum Sample Size p for

Proportion

0.90

Confidence Level 0

Sample Proportion 0.5

Error 0.04

z-Value 1.64

5

Minimum Sample

Size

423

51. Virginia wants to estimate the percentage of students who live more than three miles from the school. She wants to create a 95% confidence interval which has an error bound of at most 5%. How many students should be polled to create the confidence interval?

Answer: 385

Minimum Sample Size p for

Proportion

Confidence Level 0.950

Sample Proportion 0.5

Error 0.05

z-Value 1.960

Minimum Sample

Size

385

52. Suppose an automotive repair company wants to determine the current percentage of customers who keep up with regular vehicle maintenance. How many customers should the company survey in order to be 95% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion of customers who keep up with regular vehicle maintenance?

Answer: 601

Minimum Sample Size p for

Proportion

0.95

Confidence Level 0

Sample Proportion 0.5

Error 0.04

z-Value 1.96

0

Minimum Sample

Size

601

53. Suppose a clothing store wants to determine the current percentage of customers who are over the age of forty. How many customers should the company survey in order to be 92% confident that the estimated (sample) proportion is within 5percentage points of the true population proportion of customers who are over the age of forty?

Answer: 307

Minimum Sample Size p for

Proportion

0.92

Confidence Level 0

Sample Proportion 0.5

Error 0.05

1.75

z-Value 1

Minimum Sample Size 307

54. The average height of a population is unknown. A random sample from the population yields a sample mean of x¯=66.3inches. Assume the sampling distribution of the mean has a standard deviation of σx¯=0.8 inches. Use the Empirical Rule to construct a 95% confidence interval for the true population mean height. Provide your answer below: 64.7 – 67.9

55. In a random sample of 30 young bears, the average weight at the age of breeding is 312 pounds. Assuming the population ages are normally distributed with a population standard deviation is 30 pounds, use the Empirical Rule to construct a 68%confidence interval for the population average of young bears at the age of breeding. Do not round intermediate calculations. Round only the final answer to the nearest pound. Remember to enter the smaller value first, then the larger number.

Answer: 307 – 317

56. In a food questionnaire, a random sample of teenagers were asked whether they like pineapple pizza. The questionnaire resulted in a sample proportion of p′=0.43, with a sampling standard deviation of σp′=0.06, who like this type of pizza. Write a 99.7% confidence interval using the Empirical Rule for the true proportion of teenagers who like pineapple pizza.

Answer: 0.25 – 0.61

57. A marine biologist is interested in whether the Chinook salmon, a particular species of salmon in the Pacific Northwest, are getting smaller within the last decade. In a random sample of this species of salmon, she found the mean length was x¯=36inches with a margin of error of 9 inches. Construct a confidence interval for the mean length of Chinook salmon.

Answer: 27 – 45

58. A researcher is trying to estimate the population mean for a certain set of data. The sample mean is 45, and the error bound for the mean is 9, at a 99.7% confidence level. (So, x¯=45 and EBM = 9.) Find and interpret the confidence interval estimate.

Answer: We can estimate, with 99.7% confidence that the true value of the population means is between 36 and 54.

59. A random sample of registered voters were asked about an issue on the ballot of an upcoming election. The proportion of those surveyed who plan to vote " Yes" on the issue is 0.54, with a margin of error of 0.06. Construct a confidence interval for the proportion of registered voters that plan to vote " Yes" on the issue.

Answer: .48 – .60

(0.54-0.06 ; 0.54 + 0.06)

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